Gujarat Board Statistics Class 11 GSEB Solutions Chapter 4 Measures of Dispersion Ex 4.1 Textbook Exercise Questions and Answers.
Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.1
Question 1.
The following data refer to the heights in cms. of 10 students of a class. Find range and coefficient of range of height of the students.
Answer:
Here, xH = 185 cm; xL=145cm
Range of height:
R = xH – xL
= 185 – 145
= 40 cm
Coefficient of range = \(\frac{x_{\mathrm{H}}-x_{\mathrm{L}}}{x_{\mathrm{H}}+x_{\mathrm{L}}}\)
= \(\frac{185-145}{185+145}\)
= \(\frac{40}{330}\)
= 0.12
Question 2.
A bus company has 77 buses for travelling in the city. The information of number of passengers in bus on a particular day at a particular time is given below. Find the range and coefficient of range of number of passengers.
Answer:
Here. XH = Maximum passengers travelling in a bus = 37
XL = Minimum passengers travelling in a bus = 2
Range of number of passengers:
R = xH – xL
= 37 – 2
= 35 Passengers
Coefficient of range = \(\frac{\mathrm{R}}{x_{\mathrm{H}}+x_{\mathrm{L}}}\)
= \(\frac{35}{37+2}\)
= \(\frac{35}{39}\)
= 0.90
Question 3.
Using the following frequency distribution of marks of students of a school, find range and relative range of the marks:
Answer:
Here, xH = Upper limit of the last class 70 – 80 = 80 marks
xL = Lower limit of the initial class 20 – 30 = 20 marks
Range of marks of students :
R = xH – xL
= 80 – 20 = 60 marks
Relative range of marks:
Relative range = \(\frac{\mathrm{R}}{x_{\mathrm{H}}+x_{\mathrm{L}}}\)
= \(\frac{60}{80+20}\)
= \(\frac{60}{100}\)
= 0.60
Question 4.
The frequency distribution of daily income , (in thousand of 80 shops of an area is as follows. Find the absolute and the relative measure of range of daily income from it.
Answer:
Here, xH = Upper limit of the last class 30 – 34 = ₹ 34
xL = Lower limit of the initial class 5-9 = ₹ 5
Absolute measure of range of age:
R = xH – xL
= 34 – 5 = ₹ 29 thousand
Relative measure of range of age:
Coefficient of range = \(\frac{x_{\mathrm{H}}-x_{\mathrm{L}}}{x_{\mathrm{H}}+x_{\mathrm{L}}}=\frac{\mathrm{R}}{x_{\mathrm{H}}+x_{\mathrm{L}}}\)
= \(\frac{29}{34+5}\)
= \(\frac{29}{39}\)
= 0.74